Chemistry: Atoms First

Presentation on theme: "Chemistry: Atoms First"— Presentation transcript:

Chemistry: Atoms First
Julia Burdge & Jason Overby Chapter 8 Chemical Reactions Kent L. McCorkle Cosumnes River College Sacramento, CA

8 Stoichiometry: Ratios of Combination 8.1 Chemical Equations
Interpreting and Writing Chemical Equations Balancing Chemical Equations Patterns of Chemical Reactivity 8.2 Combustion Analysis Determination of Empirical Formula 8.3 Calculations with Balanced Chemical Equations Moles of Reactants and Products Mass of Reactants and Products 8.4 Limiting Reactants Determining the Limiting Reactant Reaction Yield 8.5 Periodic Trends in Reactivity of the Main Group Elements General Trends in Reactivity Reactions of the Active Metals Reactions of the Other Main Group Elements Comparison of Group 1A and Group 1B Elements

Chemical Equations 8.1 A chemical equation uses chemical symbols to denote what occurs in a chemical reaction. NH3 + HCl → NH4Cl Ammonia and hydrogen chloride react to produce ammonium chloride. Each chemical species that appears to the left of the arrow is called a reactant. Each species that appears to the right of the arrow is called a product.

NH3(g) + HCl(g) → NH4Cl(s)
Interpreting and Writing Chemical Equations Labels are used to indicate the physical state: (g) gas (l) liquid (s) solid (aq) aqueous [dissolved in water] NH3(g) + HCl(g) → NH4Cl(s) SO3(g) + H2O(l) → H2SO4(aq)

Balancing Chemical Equations
Chemical equations must be balanced so that the law of conservation of mass is obeyed. Balancing is achieved by writing stoichiometric coefficients to the left of the chemical formulas.

Balancing Chemical Equations
Generally, it will facilitate the balancing process if you do the following: Change the coefficients of compounds before changing the coefficients of elements. Treat polyatomic ions that appear on both sides of the equation as units. Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient.

Balancing Chemical Equations
Write the balanced chemical equation that represents the combustion of propane. Solution: Step 1: Write the unbalanced equation: C3H8(g) + O2(g) → CO2(g) + H2O(l) Step 2: Leaving O2 until the end, balance each of the atoms: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) Step 3: Double check to make sure there are equal numbers of each type on atom on both sides of the equation.

Ba(OH)2(aq) + HClO4(aq) → Ba(ClO4)2(aq) + H2O(l)
Worked Example 8.1 Write and balance the chemical equation for the aqueous reaction of barium hydroxide and perchloric acid to produce aqueous barium perchlorate and water. Strategy The reactants are Ba(OH)2 and HClO4, and the products are Ba(ClO4)2 and H2O. Because the reaction is aqueous, all species except H2O will be labeled (aq) in the equation. Being a liquid, H2O will be labeled (l). Adjust the coefficients to ensure that there are identical numbers of each type of atom on both sides of the reaction arrow. Solution The chemical statement “barium hydroxide and perchloric acid react to produce barium perchlorate and water” can be represented with the following unbalanced equation: Ba(OH)2(aq) + HClO4(aq) → Ba(ClO4)2(aq) + H2O(l) Perchlorate ions (ClO4-) appear on both sides of the equation, so count them as units, rather than count the individual atoms they contain. Thus, the tally of atoms and polyatomic ions is

Worked Example 8.1 (cont.) Solution (cont.)
Ba(OH)2(aq) + HClO4(aq) → Ba(ClO4)2(aq) + H2O(l) 1 – Ba – 1 2 – O – 1 3 – H – 2 1 – ClO4- – 2 The barium atoms are already balanced, and placing a coefficient of 2 in front of HClO4(aq) balances the number of perchlorate ions. Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + H2O(l) 4 – H – 2 2 – ClO4- – 2 (not including O atoms in ClO4- ions) (not including O atoms in ClO4- ions)

Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l)
Worked Example 8.1 (cont.) Solution (cont.) Placing a 2 in front of H2O(l) balances both the O and H atoms, giving us the final balanced equation: Ba(OH)2(aq) + 2HClO4(aq) → Ba(ClO4)2(aq) + 2H2O(l) 1 – Ba – 1 2 – O – 2 4 – H – 4 2 – ClO4- – 2 (not including O atoms in ClO4- ions) Think About It Check to be sure the equation is balanced by counting all the atoms individually. 1 – Ba – 1 10 – O – 10 4 – H – 4 2 – Cl – 2

Worked Example 8.2 Butyric acid (also known as butanoic acid, C4H8O2) is one of many compounds found in milk fat. First isolated from rancid butter in 1869, burtyic acid has received a great deal of attention in recent years as a potential anticancer agent. Write a balanced equation for the metabolism of butyric acid. Assume that the overall process of metabolism and combustion are the same (i.e., reaction with oxygen to produce carbon dioxide and water). Strategy Begin by writing an unbalanced equation to represent the combination of reactants and formation of products as stated in the problem, and then balance the equation.

Worked Example 8.2 (cont.) Solution
C4H8O2(aq) + O2(g) → CO2(g) + H2O(l) Balance the number of C atoms by changing the coefficient for CO2 from 1 to 4. C4H8O2(aq) + O2(g) → 4CO2(g) + H2O(l) Balance the number of H atoms by changing the coefficient for H2O from 1 to 4. C4H8O2(aq) + O2(g) → 4CO2(g) + 4H2O(l) Finally, balance the number of O atoms by changing the coefficient for O2 from 1 to 5. C4H8O2(aq) + 5O2(g) → 4CO2(g) + 4H2O(l) Think About It Count the number of each type of atom on each side of the reaction arrow to verify that the equation is properly balanced. The are 4 C, 8 H, and 12 O in the reactants and in the products, so the equations is balanced.

Patterns of Chemical Reactivity
Three of the most commonly encountered reaction types are combination, decomposition, and combustion. Combination – two or more reactants combine to form a single product NH3(g) + HCl(g) → NH4Cl(s) Decomposition – two or more products form from a single reactant CaCO3(s) → CaO(s) + CO2(g) Combustion – a substance burns in the presence of oxygen. Combustion of a compound that contains C and H (or C, H, and O) produces carbon dioxide gas and water. CH2O(l) + O2(g) → CO2(g) + H2O(l) Δ

Worked Example 8.3 Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: (a) H2(g) + Br2(g) → 2HBr(g), (b) 2HCO2H(l) + O2(g) → 2CO2(g) + 2H2O(l), (c) 2KClO3(s) → 2KCl(s) + 3O2(g). Strategy The equation in part (a) depicts two reactants and one product. The equation in part (b) represents a combination of a compound containing C, H, and O–with O2–to produce CO2 and H2O. The equation in part (c) represents two products being formed from a single reactant. Solution These equations represent (a) a combination reaction, (b) a combustion reaction, and (c) a decomposition reaction. Think About It Make sure that a reaction identified as combination has only one product [as in part (a)], a reaction identified as combustion consumes O2 and produces CO2 and H2O [as in part (b)], and a reaction identified as a decomposition has only one reactant [as in part (c)].

Combustion Analysis 8.2 The experimental determination of an empirical formula is carried out by combustion analysis.

Combustion Analysis In the combustion of 18.8 g of glucose, 27.6 g of CO2 and 11.3 g of H2O are produced. It is possible to determine the mass of carbon and hydrogen in the original sample as follows: The remaining mass is oxygen: 18.8 g glucose – (7.53 g C g H) = 10.0 g O

Combustion Analysis It is now possible to calculate the empirical formula. Step 1: Determine the number of moles of each element. Step 2: Write the empirical formula and divide by the smallest subscript to find the whole number ratio. C0.627H1.25O0.626 simplifies to CH2O

Combustion Analysis The molecular formula may be
determined from the empirical formula if the approximate molecular mass is known. To determine the molecular formula, divide the molar mass by the empirical formula mass. For glucose: Empirical formula: CH2O Empirical formula mass: [12.01 g/mol + 2(1.008 g/mol) g/mol] ≈ 30 g/mol Molecular mass: 180 g/mol Molecular mass/Empirical mass: 180/30 = 6 Molecular formula = [CH2O] x 6 = C6H12O6

Worked Example 8.4 Combustion of a 5.50-g sample of benzene produces g CO2 and 3.81 g H2O. Determine the empirical formula and the molecular formula of benzene, given that its molar mass is approximately 78 g/mol. Strategy Determine the mass of C and H in the 5.50-g sample of benzene. Sum these masses and subtract from the original sample mass to find the mass of O. Convert the mass of each sample to moles, and use the results as subscripts in a chemical formula. Convert the subscripts to whole numbers by dividing each number by the smallest subscript to obtain the empirical formula. To calculate the molecular formula, first divide the molar mass given in the problem by the empirical formula mass. Then, multiply the subscripts in the empirical formula by the resulting number to obtain the subscripts in the molecular formula.

Worked Example 8.4 (cont.) Solution We calculate the mass of carbon and the mass of hydrogen in the products (and therefore the original 5.50-g sample) as follows: The total mass of products is g g = g. Because the combined masses of C and H account for the entire mass of the original sample (5.499 ≈ 5.50 g), this compound must not contain O. 1 mol CO2 44.01 g CO2 1 mol C 1 mol CO2 12.01 g C 1 mol C mass of C = g CO2 × × × = g C 1 mol H2O 44.01 g H2O 1 mol H 1 mol H2O 1.008 g H 1 mol H mass of H = 3.81 g H2O × × × = g C

Worked Example 8.4 (cont.) Solution Converting mass to moles for each element present in the compound, gives the formula C0.4244H Converting the subscripts to whole numbers (0.4224/ ≈ 1; 0.423/ ≈ 1) gives the empirical formula CH. Finally, dividing the approximate molar mass (78 g/mol) by the empirical molar mass (12.01 g/mol g/mol = g/mol) gives 78/13.02 ≈ 6. Then, multiplying both subscripts in the empirical formula by 6 gives the molecular formula C6H6. 1 mol C 12.01 g C moles of C = g C × = mol C 1 mol H 12.01 g H moles of H = 3.81 g H × = mol H Think About It Use the molecular formula to determine the molar mass and make sure that the result agrees with the molar mass given in the problem. For C6H6, the molar mass is 6(12.01 g/mol) + 6(1.008 g/mol) = g/mol, which agrees with the 78 g/mol given in the problem statement.

8.3 Calculations with Balanced Chemical Equations
Balanced chemical equations are used to predict how much product will form from a given amount of reactant. 2 moles of CO combine with 1 mole of O2 to produce 2 moles of CO2. 2 moles of CO is stoichiometrically equivalent to 2 moles of CO2.

Calculations with Balanced Chemical Equations
Consider the complete reaction of 3.82 moles of CO to form CO2. Calculate the number of moles of CO2 produced.

Calculations with Balanced Chemical Equations
Consider the complete reaction of 3.82 moles of CO to form CO2. Calculate the number of moles of O2 needed.

Worked Example 8.5 Urea [(NH2)2CO] is a by-product of protein metabolism. This waste product is formed in the liver and then filtered from the blood and excreted in the urine by the kidneys. Urea can be synthesized in the laboratory by the combination of ammonia and carbon dioxide according to the equation (a) Calculate the amount of urea that will be produced by the complete reaction of 5.25 moles of ammonia. (b) Determine the stoichiometric amount of carbon dioxide required to react with 5.25 moles of ammonia. Strategy Use the balanced chemical equation to determine the correct stoichiometric conversion factors, and then multiply by the number of moles of ammonia given.

Worked Example 8.5 (cont.) Solution
(a) moles (NH2)2CO produced = 5.25 mol NH3 × (b) moles CO2 produced = 5.25 mol NH3 × 1 mol (NH2)2CO 2 mol NH3 = 2.63 mol (NH2)2CO 1 mol CO2 2 mol NH3 = 2.63 mol CO2 Think About It As always, check to be sure that units cancel properly in the calculation. Also, the balanced equation indicates that there will be fewer moles of urea produced than ammonia consumed. Therefore, your calculated number of moles of urea (2.63) should be smaller than the number of moles given in the problem (5.25). Similarly, the stoichiometric coefficients in the balanced equation are the same for carbon dioxide and urea, so your answers to this problem should also be the same for both species.

Worked Example 8.6 Dinitrogen monoxide (N2O), also known as nitrous oxide or “laughing gas,” is used as an anesthetic in dentistry. It is manufactured by heating ammonium nitrate. The balanced equation is NH4NO3(s) → N2O(g) + 2H2O(l) (a) Calculate the mass of ammonium nitrate that must be heated in order to produce 10.0 g of nitrous oxide. (b) Determine the corresponding mass of water produced in the reaction. Δ Strategy For part (a), use the molar mass of nitrous oxide to convert the given mass of nitrous oxide to moles, use the appropriate stoichiometric conversion factor to convert to moles of ammonium nitrate, and then use the molar mass of ammonium nitrate to convert to grams of ammonium nitrate. For part (b), use the molar mass of nitrous oxide to convert the given mass of nitrous oxide to moles, use the stoichiometric conversion factor to convert from moles of nitrous oxide to moles of water, and then use the molar mass of water to convert to grams of water.

Worked Example 8.6 (cont.) Solution (a) 10.0 g N2O × 1 mol N2O
0.227 mol NH4NO3 × Thus, 18.2 g of ammonium nitrate must be heated in order to produce 10.0 g of nitrous oxide. (b) Starting with the number of moles of nitrous oxide determined in the first step of (a), 0.454 mol H2O × 1 mol N2O 44.02 g N2O = mol N2O 1 mol NH4NO3 1 mol N2O Think About It Use the law of conservation of mass to check your answers. Make sure that the combined mass of both products is equal to the mass of reactant you determined in part (a). In this case (rounded to the appropriate number of significant figures), 10.0 g g = 18.2 g. Remember that small differences may arise as the result of rounding. = mol NH4NO3 80.05 g NH4NO3 1 mol NH4NO3 = 18.2 g NH4NO3 2 mol H2O 1 mol N2O = mol H2O 18.02 g H2O 1 mol H2O = 8.18 g H2O

CO(g) + 2H2(g) → CH3OH(l) 8.4 Limiting Reactants
The reactant used up first in a reaction is called the limiting reactant. Excess reactants are those present in quantities greater than necessary to react with the quantity of the limiting reactant. CO(g) + 2H2(g) → CH3OH(l)

Limiting Reactants Consider the reaction between 5 moles of CO and 8 moles of H2 to produce methanol. How many moles of H2 are necessary in order for all the CO to react? How many moles of CO are necessary in order for all of the H2 to react? 10 moles of H2 required; 8 moles of H2 available; limiting reactant. 4 moles of CO required; 5 moles of CO available; excess reactant. CO(g) + 2H2(g) → CH3OH(l)

3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
Worked Example 8.7 Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7) react to form carbon dioxide gas, among other products. 3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq) The formation of CO2 causes the trademark fizzing when the tablets are dropped into a glass of water. An Alka-Seltzer tablet contains g of sodium bicarbonate and g citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO2 forms. Strategy Convert each of the reactant masses to moles. Use the balanced equation to write the stoichiometric conversion factor and determine which reactant is limiting. Next, determine the number of moles of excess reactant remaining and the number of moles of CO2 produced. Finally, use the appropriate molar masses to convert moles of excess reactant and moles of CO2 to grams.

Worked Example 8.7 (cont.) Solution 1.700 g NaHCO3 × 1 mol NaHCO3
1.000 g H3C6H5O7 × (a) To determine which reactant is limiting, calculate the amount of citric acid necessary to react completely with mol sodium bicarbonate. mol NaHCO3× The amount of H3C6H5O7 required to react with mol of NaHCO3 is more than a tablet contains. Therefore, citric acid is the limiting reactant and sodium bicarbonate is the excess reactant. 1 mol NaHCO3 84.01 g NaHCO3 = mol NaHCO3 1 mol H3C6H5O7 g H3C6H5O7 = mol H3C6H5O7 1 mol H3C6H5O7 3 mol NaHCO3 = mol H3C6H5O7

Worked Example 8.7 (cont.) Solution
(b) To determine the mass of excess reactant (NaHCO3) left over, first calculate the amount of NaHCO3 that will react: mol H3C6H5O7 × Thus, mole of NaHCO3 will be consumed, leaving mole unreacted. Convert the unreacted amount to grams as follows: mol NaHCO3 × 3 mol NaHCO3 1 mol H3C6H5O7 = mol NaHCO3 84.01 g NaHCO3 1 mol NaHCO3 = g NaHCO3

Worked Example 8.7 (cont.) Solution
(c) To determine the mass of CO2 produced, first calculate the moles of CO2 produced from the number of moles of limiting reactant (H3C6H5O7) consumed. mol H3C6H5O7 × Convert this amount to grams as follows: mol CO2× To summarize the results: (a) citric acid is the limiting reactant, (b) g sodium bicarbonate remains unreacted, and (c) g carbon dioxide is produced. Think About It In a problem such as this, it is a good idea to check your work by calculating the amounts of the other products in the reaction. According to the law of conservation of mass, the combined starting mass of the two reactants (1.700 g g = g) should equal the sum of the masses of products and leftover excess reactant. In this case, the masses of H2O and Na3C6H5O7 produced are g and g, respectively. The mass of CO2 produced is g [from part (c)] and the amount of excess NaHCO3 is g [from part (b)]. The total, g g g g, is g, identical to the total mass of reactants. 3 mol CO2 1 mol H3C6H5O7 = mol CO2 44.01 g CO2 1 mol CO2 = g CO2

Limiting Reactants The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product. The actual yield is the amount of product actually obtained from a reaction. The percent yield tells what percentage the actual yield is of the theoretical yield.

Worked Example 8.8 Aspirin, acetylsalicylic acid (C9H8O4), is the most commonly used pain reliever in the world. It is produced by the reaction of salicylic acid (C7H6O3) and acetic anhydride (C4H6O3) according to the following equation: In a certain aspirin synthesis, g of salicylic acid and g of acetic anhydride are combined. Calculate the percent yield if g of aspirin are produced. C7H6O3 salicylic acid + C4H6O3 acetic anhydride C9H8O4 acetylsalicylic acid + HC2H3O2 acetic acid Strategy Convert reactant grams to moles, and determine which is the limiting reactant. Use the balanced equation to determine the moles of aspirin that can be produced and convert to grams for the theoretical yield. Use this and the actual yield given to calculate the percent yield.

Worked Example 8.8 (cont.) Solution 104.8 g C7H6O3× 1 mol C7H6O3
Because the two reactants combine in a 1:1 mole ratio, the reactant present in the smallest number of moles (in this case, salicylic acid) is the limiting reactant. According to the balanced equation, one mole of aspirin is produced for every mole of salicylic acid consumed. Therefore, the theoretical yield of aspirin is mol. We convert this to grams using the molar mass of aspirin: mol C9H8O4× 1 mol C7H6O3 g C7H6O3 = mol C7H6O3 1 mol C4H6O3 g C4H6O3 = mol C4H6O3 g C9H8O4 1 mol C9H8O4 = g C9H8O4

Worked Example 8.8 (cont.) Solution Thus, the theoretical yield is g. If the actual yield is 105.6, the percent yield is % yield = 105.6 g 136.7 g ×100% = 77.25% Think About It Make sure you have used the proper molar masses and remember that percent yield can never exceed 100 percent.

8.5 Periodic Trends in Chemical Properties of the Main Group Elements
Ionization Energy and Electron Affinity enable us to understand types of reactions that elements undergo and the types of compounds formed.

CaH2(s) + H2O(l) → Ca(OH)2(aq) + H2(g)
General Trends in Reactivity Hydrogen (1s1) Grouped by itself Forms a cation with a +1 charge (H+) Forms an anion with a -1 charge (H-) Hydrides react with water to produce hydrogen gas and a base. CaH2(s) + H2O(l) → Ca(OH)2(aq) + H2(g)

Reactions of the Active Metals
Group 1A Elements (ns1, n ≥ 2) Low IE Never found in nature in pure elemental state React with oxygen to form metal oxides sodium potassium cesium Alkali metals reacting with water

Barium reacting with water
Reactions of the Active Metals Group 2A Elements (ns2, n ≥ 2) Less reactive than 1A Some react with H2O to produce H2 Some react with acid to produce H2 Barium reacting with water

Finely-divided aluminum sprinkled into a flame to form Al2O3
Reactions of Other Main Group Elements Finely-divided aluminum sprinkled into a flame to form Al2O3 Group 3A elements (ns2np1, n ≥ 2) Metalloid (B) and metals (all others) Al forms Al2O3 with oxygen Al forms +3 ions in acid Others form +1 and +3

Reactions of Other Main Group Elements
Group 4A elements (ns2np2, n ≥ 2) Nonmetal (C); metalloids (Si, Ge) and others metals Form +2 and +4 oxidation states Sn, Pb react with acid to produce H2

Reactions of Other Main Group Elements
Group 5A elements (ns2np3, n ≥ 2) Nonmetal (N2, P), metalloid (As, Sb), and metal (Bi) Nitrogen, N2, forms variety of oxides Phosphorus, P4 As, Sb, Bi (crystalline) HNO3 and H3PO4 important industrially

Reactions of Other Main Group Elements
Nonmetal oxides added to water produce an acid Group 6A elements (ns2np4, n ≥ 2) Nonmetals (O, S, Se) Oxygen, O2 Sulfur, S8 Selenium, Se8 Metalloids (Te, Po) Te, Po (crystalline) SO2, SO3, H2S, H2SO4 Forest damaged by acid rain

Reactions of Other Main Group Elements
Group 7A elements (ns2np5, n ≥ 2) All diatomic Do not exist in elemental form in nature Form ionic “salts” Form molecular compounds with each other React with hydrogen to form hydrogen halides

Reactions of Other Main Group Elements
Group 8A elements (ns2np6, n ≥ 2) All monatomic Filled valence shells Considered “inert” until 1963 when Xe and Kr were used to form compounds No major commercial use

Comparison of Group 1A and Group 1B Elements
Have single valence electron Properties differ Group 1B much less reactive than 1A High IE of 1B - incomplete shielding of nucleus by inner “d” -outer “s” electron of 1B strongly attracted to nucleus 1B metals often found elemental in nature (coinage metals)

8 Chapter Summary: Key Points Chemical Equations
Interpreting and Writing Chemical Equations Balancing Chemical Equations Patterns of Chemical Reactivity Combustion Analysis Determination of Empirical Formula Calculations with Balanced Chemical Equations Moles of Reactants and Products Mass of Reactants and Products Limiting Reactants Determining the Limiting Reactant Reaction Yield Periodic Trends in Reactivity of the Main Group Elements General Trends in Reactivity Reactions of the Active Metals Reactions of the Other Main Group Elements Comparison of Group 1A and Group 1B Elements

Balance the following equations:
Group Quiz #14 Balance the following equations: __Fe2(SO4)3 + __K3PO4  __K2SO4 + __FePO4 __Hg(NO3)2 + __ KI  __HgI2 + __ KNO3 __Li2O(s) + __ H2O  __LiOH __HBr + __Ba(OH)2  __BaBr2 + __H2O __K2PtCl4 + __NH3  __Pt(NH3)2Cl2 + __KCl 51

How many atoms are in 12.987 g of iron?
Group Quiz #15 How many atoms are in g of iron? How many H atoms are in a gram sample of H2O? 52

Group Quiz #16 Complete the equation below (with phases) and balance it. K2S(aq) + AgNO3(g)  If you combine mL of M K2S with g AgNO3, how much solid product can be formed? If g of product were actually formed, what is the percent yield?